Question

A tube of length 1m is filled completely with an ideal liquid of mass 2M, and closed at both ends. The tube is rotated uniformly in horizontal plane about one of its ends. If the force exerted by the liquid at the other end is \( F \) and the angular velocity of the tube is \( \omega \), then the value of \( \alpha \) is ______ in SI units.

Hint:

For rotational motion, remember that the centripetal force depends on the mass, the radius of rotation, and the square of the angular velocity.

Answer 1

Correct Answer: 1

To determine the liquid's force at the rotating tube's far end, we apply the centrifugal force concept. Given:

• Tube length \( L = 1 \) m
• Liquid mass \( m = 2M \)
• Angular velocity \( \omega \)

The centrifugal force \( dF \) at distance \( x \) from the pivot is \( dF = \omega^2 \cdot x \cdot dm \). With a mass density \( \rho = \frac{2M}{L} = \frac{2M}{1} \), a small segment's mass is \( dm = \rho \cdot dx = 2M \cdot dx \). Substituting this into the force equation yields \( dF = \omega^2 \cdot x \cdot (2M \cdot dx) = 2M \omega^2 x \, dx \). The total force \( F \) at the tube's end is the integral of \( dF \) from \( x=0 \) to \( x=1 \):
\( F = \int_0^1 2M \omega^2 x \, dx = 2M \omega^2 \left[\frac{x^2}{2}\right]_0^1 \)
\( F = 2M \omega^2 \cdot \frac{1^2}{2} = M \omega^2 \).
The coefficient of \( M \omega^2 \) in the force expression is \( \alpha \). Thus, \( \alpha = 1 \). This result falls within the expected range of 1 to 1, validating the calculation.